for eq. of plane, does the normal vector's "tip" (end point) have to be the same point as the point when they

Question

say "the plane through the point P(x0y0z0)", so basically, does P(x0y0z0) line on the normal vector?

Answer 1

no, not exactly. the normal vector gives you a DIRECTION (like slope in 2 dimensions). you can go that direction from any point. the equation of a line in n-space is just a point, P, plus multiples of a direction vector, tV. it says you can get to any point on the line if you start at this point (P)and go a sufficient distance in this direction (tV).

you get the equation of a plane through a given point by saying that the dot product of the vector from the given point to any point in the plane, and the normal vector, is 0.

in the xy plane, there's an infinite number of lines with the same slope. so in n-space, a vector is just a direction. to say the point (a,b,c) is the vector's "tip" assumes the vector's starting point is (0,0,0). but the vector can start anywhere.

Answer 2

A plane's normal vector just gives a direction. It is not 'nailed down' . You can conceptually move the vector perpindicular to its directrion. It's like a pole that stick straight up. You can move the pole anywhere around on the flat surface and the pole (vector) is the same.

Answer 3

No, it may't be. We already understand the universe is continuing to enhance... even extra-so, all of us understand that that fee of enlargement is once back accelerating, and if pink shift measurements are astounding, the outer radius of the universe is already accelerating quicker than mild velocity, and the radius of superluminal enlargement will slowly shrink. As such, there should not be a large crunch. ----------- "There are 2 theories for the top of the universe:" No, there are 2 hypotheses. One has already been shown to be quite not likely, even no longer achieveable in accordance with attainable observations. considering gravity promulgates on the speed of sunshine, if there are aspects of the universe already receeding at superluminal velocity, then the gravitational results of those aspects of the universe can't in any respect attain something else of the universe -- as such, they have been removed from the gravitational properly. this suggests the classic gravitational density of the universe is lowering critically with each and each passing second because the lightcone of any given element contains a lot less and a lot less mass. the classic density of the universe is measured as being quite on the fringe of one, if the gravitational density of the universe critically decreases, then the universe is open and there can't be a large crunch problem.

Answer 4

no its not necessary that the point P lies on the normal vector,until and unless not mentioned in the question.